STRUCTURE OF Ho-165
By Prof Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of electromagnetic laws in favour of wrong theories which could not lead to the nuclear structure. Under this physics crisis and using the charged Up and Down quarks, discovered by Gell-Mann and Zweig, in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the nuclear binding and the correct nuclear structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). ' ' STRUCTURES OF HOLMIUM Natural holmium (Ho) contains one stable isotope, Ho-165. A number of synthetic radioactive isotopes are known, the most stable one is Ho-163, with a half-life of 4,570 years. Comparing the structures of holmium of 67 protons (odd number) with those of Dysprosium of 66 protons (even number) we see that the nuclides of holmium break the high symmetry of Dysprosium . (See my STRUCTURE OF Dy-156...Dy-164 ). After a careful analysis of this comparison I discovered that the additional vertical p67n67 forms with the n65p65 the alpha particle in front of the central parallelepiped giving 2(n) of opposite spins. Then for symmetrical arrangements the n38p38 is moved from the up square to the p66n66 for making the symmetrical alpha particle behind the central parallelepiped which gives also 2(n) of opposite spins. Moreover for making a stable structure the deuteron n40p40 of the up square makes with p17 and n19 a vertical rectangle. In this case the elongated shape of two horizontal squares becomes a non elongated shape in order to avoid the strong pp repulsions of the elongation. Especially the nucleons of the six horizontal planes give S =0 while the down square gives 4n of negative spins which contribute to the total spin S =-7/2 of the Ho-165. So under these arrangements the number N of blank positions is given by The down square gives 4n with negative spins. The first and the sixth plane give 4(n) of weak bonds with opposite spins. The second and the fifth plane give 4{n} with three bonds per neutron and 8n of opposite spins. The third and the fourth plane give 4(n) of weak bonds with opposite spins Also the four alpha particles give at the third and the fourth plane 8(n) of opposite spins. That is N = 4{n} +12n + 16(n) = 32 blank positions able to receive 18 extra neutrons of negative spins and 14 extra neutrons of positive spins. ' ' STRUCTURE OF Ho-165 WITH S = -7/2 Since the Ho-165 of 31 extra neutrons has S = -7/2 we conclude that it is due to the summation of S =-2 of the two deuterons like p37n37 and n39p39 and the total spin S =-3/2 of the 31 extra neutrons . In other words it has 17 extra neutrons of negative spins and 14 extra neutrons of positive spins. Note that the unstable Ho-167 of 33 extra neutrons with the same S =-7/2 has 18 extra neutrons of negative spins and 15 extra neutrons of positive spins. Since 15>14 we see that it has one neutron of positive spin creating a single bond which leads to the beta decay. ' ' ' DIAGRAM OF HOLMIUM FORMING 32 BLANK POSITIONS' The additional p67 and n67, the symmetrical n38p38 and the n40p40 are not shown. But you can see the p67 and n38 by using the top view of the third plane. Note that the p47n47 along with the p48n48 make in the interior shapes the two symmetrical alpha particles of opposite spins . But you cannot see the p49n49, the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' ' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth h. plane' ' n..........p29.........n10.........p10…… n30 ' ' n29…..p9..........n9 …….p30.........n Fifth h. plane' ' n61....p47..........n27.........p8..........n8.........p28........... n48......p62' ' p39....n45...........p27........n7.........p7........n28..........p46...........n63 Fourth h. plane ' ' p61......n47..........p25.........n6.........p6..........n26...........p48.....n62' ' 'n39....p45..........n25……p5..........n5……….p26.......n46 .........p63 Third h. plane ' n23………p4........n4………….p24..............n' ' n......p23…….....n3……….p3………..n24 Second h. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] First h. plane' ' n.........p37......n37 ' ' n39......p39 Down Square with n' TOP VIEW OF THE FIRST HORIZONTAL PLANE '''Here you see the 2(n) of weak horizontal bonds ' ' (n)........p34....... n34''' ' p21.......n2........ p2.......n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24..........n ' ' n.......p23........n3........p3.........n24' ' {n}...... p13......n13 ' ' n' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' Here the p66 and n38 are shown near the n52 and p51 respeciively. Thus all alpha particles of the third and the fourth horizontal plane give 8(n). Using this top view of the third plane you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. ' ' ' (n) ' p66......n38 ' ' (n)........p58....... n50.......p51....n60 ' ' (n) p53........n42........p16......n16......p44.........n54' ' p61 n47........p25........n6........p6........n26.........p48 n62' ' n64 p45........n25........p5........n5........p26........ n46 p63' ' n55........p41.......n15.......p15.......n43...... .p56 (n)' ' n57.......p49.......n52...... p59........(n)' n65.......p67 (n) ' ' TOP WIEW OF THE DOWN HORIZONTAL SQUARE WITH 4n OF NEGATIVE SPINS ' n' ' n......... P37......n37 ' ' n39.......p39.........n' ' n' Category:Fundamental physics concepts